Proof of the epsilon-delta form of continuity

Proof

We compare open intervals with symmetric intervals $(c - r, c + r)$.

1. Let $D \subseteq \R$, let $f \colon D \to \R$, and let $a \in D$.
2. Claim: if $V = (p, q)$ is an open interval containing a point $c$, then $r = \min(c - p, q - c)$ is positive and $(c - r, c + r) \subseteq V$.
3. Proof of the claim: $p < c < q$ gives $r > 0$. If $\abs{y - c} < r$, then $y > c - r \ge p$ and $y < c + r \le q$, so $y \in V$.
4. Assume first that $f$ is continuous at $a$. Let $\varepsilon > 0$.
5. The set $V = (f(a) - \varepsilon, f(a) + \varepsilon)$ is an open interval containing $f(a)$. The definition gives an open interval $U$ containing $a$ with $f(x) \in V$ for all $x \in U \cap D$.
6. By the claim, there is $\delta > 0$ with $(a - \delta, a + \delta) \subseteq U$.
7. Let $x \in D$ with $\abs{x - a} < \delta$. Then $x \in U \cap D$, so $f(x) \in V$, that is, $\abs{f(x) - f(a)} < \varepsilon$.
8. So the epsilon-delta condition holds.
9. Conversely, assume the epsilon-delta condition. Let $V$ be an open interval containing $f(a)$.
10. By the claim, there is $\varepsilon > 0$ with $(f(a) - \varepsilon, f(a) + \varepsilon) \subseteq V$. The condition gives $\delta > 0$ for this $\varepsilon$.
11. Let $U = (a - \delta, a + \delta)$, an open interval containing $a$.
12. Let $x \in U \cap D$. Then $\abs{x - a} < \delta$, so $\abs{f(x) - f(a)} < \varepsilon$. Hence $f(x) \in V$.
13. So $f$ is continuous at $a$ in the sense of the definition. This proves the equivalence.